[proofplan]
The proof is local near each point of the fiber. At a point $p\in F^{-1}(\{q\})$, the submersion normal form theorem gives charts in which $F$ is the coordinate projection onto $\mathbb{R}^n$. In those coordinates the fiber is exactly a coordinate slice, so the local slice charts define the embedded submanifold structure and its topology. Differentiating the projection model identifies the tangent vectors to the slice with precisely the kernel of $dF_p$.
[/proofplan]
[step:Handle the empty fiber case]
If $F^{-1}(\{q\})=\varnothing$, then the conclusion is the standard empty-manifold convention: the empty set has a unique [smooth manifold](/page/Smooth%20Manifold) structure of any prescribed dimension and is an embedded submanifold of $M$ with the [subspace topology](/page/Subspace%20Topology). The tangent-space assertion is vacuous. Hence assume for the rest of the proof that $F^{-1}(\{q\})\neq\varnothing$.
[/step]
[step:Put the submersion into local projection form near a fiber point]
Fix $p\in F^{-1}(\{q\})$. Since $F$ is a smooth submersion, the differential
\begin{align*}
dF_p:T_pM\to T_qN
\end{align*}
is surjective. By the submersion normal form theorem, there exist charts $(U,\varphi)$ on $M$ and $(V,\psi)$ on $N$ such that $p\in U$, $q\in V$, $F(U)\subset V$, $\varphi(p)=0\in\mathbb{R}^m$, $\psi(q)=0\in\mathbb{R}^n$, and the coordinate representation
\begin{align*}
\psi\circ F\circ\varphi^{-1}:\varphi(U)\to\psi(V)
\end{align*}
is the projection onto the last $n$ coordinates:
\begin{align*}
(\psi\circ F\circ\varphi^{-1})(x_1,\dots,x_m)=(x_{m-n+1},\dots,x_m).
\end{align*}
Here $m\ge n$, because $dF_p$ is a surjective [linear map](/page/Linear%20Map) from an $m$-dimensional [vector space](/page/Vector%20Space) onto an $n$-dimensional vector space.
[guided]
We first isolate the only local input needed in the proof. The hypothesis that $F$ is a submersion means that, for every point $r\in M$, the differential
\begin{align*}
dF_r:T_rM\to T_{F(r)}N
\end{align*}
is surjective. Applying this at the chosen point $p\in F^{-1}(\{q\})$ gives a surjective linear map
\begin{align*}
dF_p:T_pM\to T_qN.
\end{align*}
Surjectivity forces $\dim T_pM\ge \dim T_qN$, so $m\ge n$.
The submersion normal form theorem says that a smooth map whose differential is surjective at $p$ can be written, after choosing suitable source and target charts, as a coordinate projection. Thus we choose charts $(U,\varphi)$ around $p$ and $(V,\psi)$ around $q$ with $F(U)\subset V$, $\varphi(p)=0$, and $\psi(q)=0$, such that the coordinate representation
\begin{align*}
\psi\circ F\circ\varphi^{-1}:\varphi(U)\to\psi(V)
\end{align*}
has the form
\begin{align*}
(\psi\circ F\circ\varphi^{-1})(x_1,\dots,x_m)=(x_{m-n+1},\dots,x_m).
\end{align*}
This is the reason submersions have smooth level sets: locally, taking the fiber over $q$ becomes the elementary operation of fixing the last $n$ coordinates equal to $0$.
[/guided]
[/step]
[step:Identify the local fiber as a coordinate slice]
Define the projection map
\begin{align*}
\pi:\mathbb{R}^m\to\mathbb{R}^n,\quad \pi(x_1,\dots,x_m)=(x_{m-n+1},\dots,x_m).
\end{align*}
The preceding step says that $\psi\circ F\circ\varphi^{-1}=\pi$ on $\varphi(U)$. Define the open subset
\begin{align*}
S=\{a\in\mathbb{R}^{m-n}:(a,0)\in\varphi(U)\}\subset\mathbb{R}^{m-n},
\end{align*}
where $(a,0)$ denotes the point of $\mathbb{R}^m=\mathbb{R}^{m-n}\times\mathbb{R}^n$ with first component $a$ and second component $0\in\mathbb{R}^n$. The set $S$ is open because it is the preimage of the [open set](/page/Open%20Set) $\varphi(U)$ under the continuous inclusion
\begin{align*}
\iota:\mathbb{R}^{m-n}\to\mathbb{R}^m,\quad \iota(a)=(a,0).
\end{align*}
For $r\in U$, write $\varphi(r)=(a,b)$ with $a\in\mathbb{R}^{m-n}$ and $b\in\mathbb{R}^n$. Since $\psi(q)=0$, we have
\begin{align*}
r\in F^{-1}(\{q\})\cap U \iff \psi(F(r))=0 \iff \pi(\varphi(r))=0 \iff b=0.
\end{align*}
Therefore
\begin{align*}
\varphi(F^{-1}(\{q\})\cap U)=\varphi(U)\cap(\mathbb{R}^{m-n}\times\{0\})=S\times\{0\}.
\end{align*}
[/step]
[step:Use the slice charts to define the embedded submanifold structure]
Let
\begin{align*}
\rho:S\times\{0\}\to S,\quad \rho(a,0)=a
\end{align*}
be the coordinate projection from the slice to $\mathbb{R}^{m-n}$. Define
\begin{align*}
\theta:F^{-1}(\{q\})\cap U\to S,\quad \theta=\rho\circ\varphi|_{F^{-1}(\{q\})\cap U}.
\end{align*}
Its inverse is the map
\begin{align*}
\sigma:S\to F^{-1}(\{q\})\cap U,\quad \sigma(a)=\varphi^{-1}(a,0).
\end{align*}
Thus $\theta$ is a homeomorphism from $F^{-1}(\{q\})\cap U$, with the subspace topology, onto the open subset $S\subset\mathbb{R}^{m-n}$.
As $p$ ranges over $F^{-1}(\{q\})$, these charts cover the fiber. On overlaps, the transition maps are restrictions of smooth coordinate changes between charts of $M$ preserving the coordinate-slice condition, hence are smooth maps between open subsets of $\mathbb{R}^{m-n}$. Therefore these slice charts define a smooth structure on $F^{-1}(\{q\})$ of dimension $m-n$. In each ambient chart $(U,\varphi)$ above, the fiber is represented by the coordinate slice $S\times\{0\}$, so this smooth structure makes $F^{-1}(\{q\})$ an embedded submanifold of $M$ with the subspace topology.
[/step]
[step:Compute the tangent space of the coordinate slice]
Fix $p\in F^{-1}(\{q\})$ and use the charts from the preceding steps. Let
\begin{align*}
L:\mathbb{R}^{m-n}\to\mathbb{R}^m,\quad L(v)=(v,0)
\end{align*}
be the linear inclusion of the slice directions, and let
\begin{align*}
P:\mathbb{R}^m\to\mathbb{R}^n,\quad P(u_1,\dots,u_m)=(u_{m-n+1},\dots,u_m)
\end{align*}
be the linear projection onto the last $n$ coordinates. The coordinate expression of $dF_p$ is $P$ under the identifications
\begin{align*}
d\varphi_p:T_pM\to\mathbb{R}^m
\end{align*}
and
\begin{align*}
d\psi_q:T_qN\to\mathbb{R}^n.
\end{align*}
Indeed, differentiating $\psi\circ F\circ\varphi^{-1}=\pi$ at $0=\varphi(p)$ gives
\begin{align*}
d\psi_q\circ dF_p\circ d(\varphi^{-1})_0=P.
\end{align*}
The tangent space of the embedded coordinate slice at $p$ is carried by $d\varphi_p$ onto $\mathbb{R}^{m-n}\times\{0\}=\operatorname{im}L$. Hence, for $v\in T_pM$,
\begin{align*}
v\in T_p(F^{-1}(\{q\})) \iff d\varphi_p(v)\in\mathbb{R}^{m-n}\times\{0\}.
\end{align*}
Since $\ker P=\mathbb{R}^{m-n}\times\{0\}$ and $d\psi_q$ is a linear isomorphism, the differentiated coordinate formula gives
\begin{align*}
d\varphi_p(v)\in\ker P \iff dF_p(v)=0.
\end{align*}
Therefore
\begin{align*}
T_p(F^{-1}(\{q\}))=\ker(dF_p).
\end{align*}
[guided]
The tangent-space computation is the infinitesimal version of the slice description. The chart $\varphi$ identifies a neighbourhood of $p$ in $M$ with an open subset of $\mathbb{R}^m$, and in those coordinates the fiber is the slice
\begin{align*}
\mathbb{R}^{m-n}\times\{0\}\subset\mathbb{R}^{m-n}\times\mathbb{R}^n.
\end{align*}
Thus a tangent vector to the fiber is precisely a tangent vector whose coordinate representative points along the first $m-n$ directions and has no component in the last $n$ directions.
To make that precise, define the linear projection
\begin{align*}
P:\mathbb{R}^m\to\mathbb{R}^n,\quad P(u_1,\dots,u_m)=(u_{m-n+1},\dots,u_m).
\end{align*}
The coordinate representation of $F$ is the smooth map $\pi:\varphi(U)\to\mathbb{R}^n$ given by the same formula, so its derivative at every point is the linear map $P$. Differentiating the identity
\begin{align*}
\psi\circ F\circ\varphi^{-1}=\pi
\end{align*}
at $0=\varphi(p)$ and using the chain rule gives
\begin{align*}
d\psi_q\circ dF_p\circ d(\varphi^{-1})_0=P.
\end{align*}
Equivalently, for every $v\in T_pM$,
\begin{align*}
d\psi_q(dF_p(v))=P(d\varphi_p(v)).
\end{align*}
Now the tangent space of the coordinate slice is exactly the kernel of $P$:
\begin{align*}
\ker P=\mathbb{R}^{m-n}\times\{0\}.
\end{align*}
The embedded submanifold chart identifies $T_p(F^{-1}(\{q\}))$ with this subspace through $d\varphi_p$. Therefore
\begin{align*}
v\in T_p(F^{-1}(\{q\})) \iff d\varphi_p(v)\in\ker P.
\end{align*}
Using the differentiated coordinate formula, this is equivalent to
\begin{align*}
d\psi_q(dF_p(v))=0.
\end{align*}
Because $\psi$ is a chart, $d\psi_q:T_qN\to\mathbb{R}^n$ is a linear isomorphism, so $d\psi_q(dF_p(v))=0$ holds exactly when $dF_p(v)=0$. Hence
\begin{align*}
v\in T_p(F^{-1}(\{q\})) \iff v\in\ker(dF_p).
\end{align*}
This proves
\begin{align*}
T_p(F^{-1}(\{q\}))=\ker(dF_p).
\end{align*}
[/guided]
[/step]
[step:Conclude the global statement from the local construction]
The preceding construction applies at every point $p\in F^{-1}(\{q\})$. The resulting local slice charts cover the fiber, give it the subspace topology, and make it an embedded smooth submanifold of $M$ of dimension $m-n$. The tangent-space computation holds at each point of the fiber, so for every $p\in F^{-1}(\{q\})$,
\begin{align*}
T_p(F^{-1}(\{q\}))=\ker(dF_p).
\end{align*}
This proves the theorem.
[/step]
